(*********************************************************************** Mathematica-Compatible Notebook This notebook can be used on any computer system with Mathematica 4.0, MathReader 4.0, or any compatible application. The data for the notebook starts with the line containing stars above. To get the notebook into a Mathematica-compatible application, do one of the following: * Save the data starting with the line of stars above into a file with a name ending in .nb, then open the file inside the application; * Copy the data starting with the line of stars above to the clipboard, then use the Paste menu command inside the application. Data for notebooks contains only printable 7-bit ASCII and can be sent directly in email or through ftp in text mode. Newlines can be CR, LF or CRLF (Unix, Macintosh or MS-DOS style). NOTE: If you modify the data for this notebook not in a Mathematica- compatible application, you must delete the line below containing the word CacheID, otherwise Mathematica-compatible applications may try to use invalid cache data. For more information on notebooks and Mathematica-compatible applications, contact Wolfram Research: web: http://www.wolfram.com email: info@wolfram.com phone: +1-217-398-0700 (U.S.) Notebook reader applications are available free of charge from Wolfram Research. ***********************************************************************) (*CacheID: 232*) (*NotebookFileLineBreakTest NotebookFileLineBreakTest*) (*NotebookOptionsPosition[ 220117, 6327]*) (*NotebookOutlinePosition[ 245374, 7255]*) (* CellTagsIndexPosition[ 245330, 7251]*) (*WindowFrame->Normal*) Notebook[{ Cell[TextData[{ CounterBox["Title"], ". Precision, Interpolation, and Numerical Differentiation" }], "Title", CounterAssignments->{{"Title", 2}}], Cell[CellGroupData[{ Cell[TextData[{ CounterBox["Title"], ".", CounterBox["Section"], "\tFinite and Infinite Precision" }], "Section"], Cell[CellGroupData[{ Cell[TextData[{ CounterBox["Title"], ". ", CounterBox["Section"], ". ", CounterBox["Subsection"], "\tExact Inverse of a Hilbert Matrix" }], "Subsection"], Cell[CellGroupData[{ Cell[TextData[{ "Complete the following ", StyleBox["Table", "Input"], " command to produce an ", Cell[BoxData[ \(TraditionalForm\`n\[Times]n\)]], " Hilbert matrix, ", Cell[BoxData[ \(TraditionalForm\`\[ScriptCapitalH]\_n\)]], ", whose ", Cell[BoxData[ \(TraditionalForm\`i, j\)]], " entry is ", Cell[BoxData[ \(TraditionalForm\`1/\((i + j + 1)\)\)]], ":" }], "NumberedExercise"], Cell[BoxData[ \(TraditionalForm\`\[ScriptCapitalH]\_n_ := Table[1\/\(i + j + 1\), {i, \ n}, {j, n}]\)], "Input"] }, Open ]], Cell[CellGroupData[{ Cell[TextData[{ "Compute the determinant of ", Cell[BoxData[ \(TraditionalForm\`\[ScriptCapitalH]\_10\)]], ". What is striking about this value?" }], "NumberedExercise"], Cell[CellGroupData[{ Cell[BoxData[ FormBox[ TagBox[\(\[LeftBracketingBar]\[ScriptCapitalH]\_10\[RightBracketingBar]\ \), {Det}], TraditionalForm]], "Input"], Cell[BoxData[ \(TraditionalForm\`1\/\ 273739709893086064093902013446617579389091964235284480000000000\)], "Output"] }, Open ]], Cell["It is very small!", "NumberedAnswer"] }, Open ]], Cell[CellGroupData[{ Cell[TextData[{ "Compute the (exact) inverse of ", Cell[BoxData[ \(TraditionalForm\`\[ScriptCapitalH]\_10\)]], " and call it ", Cell[BoxData[ \(TraditionalForm\`\[ScriptCapitalI]\_10\)]] }], "NumberedExercise"], Cell[CellGroupData[{ Cell[BoxData[ \(TraditionalForm\`\[ScriptCapitalI]\_10 = Inverse[\[ScriptCapitalH]\_10]\)], "Input"], Cell[BoxData[ FormBox[ RowBox[{"(", "\[NoBreak]", GridBox[{ {"145200", \(-4247100\), "47567520", \(-277477200\), "951350400", \(-2021619600\), "2695492800", \(-2194901280\), "997682400", \(-193993800\)}, {\(-4247100\), "132509520", \(-1545944400\), "9275666400", \(-32464832400\), "70082812800", \(-94611797280\), "77819227200", \(-35667145800\), "6983776800"}, {"47567520", \(-1545944400\), "18551332800", \(-113626913400\), "404006803200", \(-883043441280\), "1204150147200", \(-998680082400\), "460929268800", \(-90789098400\)}, {\(-277477200\), "9275666400", \(-113626913400\), "707011905600", \(-2545242860160\), "5619367353600", \(-7726630111200\), "6453009763200", \(-2996040247200\), "593155442880"}, {"951350400", \(-32464832400\), "404006803200", \(-2545242860160\), "9255428582400", \(-20604346963200\), "28529095795200", \(-23968321977600\), "11185216922880", \(-2224332910800\)}, {\(-2021619600\), "70082812800", \(-883043441280\), "5619367353600", \(-20604346963200\), "46189964620800", \(-64336022150400\), "54328196482560", \(-25466342101200\), "5084189510400"}, {"2695492800", \(-94611797280\), "1204150147200", \(-7726630111200\), "28529095795200", \(-64336022150400\), "90070431010560", \(-76399026303600\), "35952482966400", \(-7202601806400\)}, {\(-2194901280\), "77819227200", \(-998680082400\), "6453009763200", \(-23968321977600\), "54328196482560", \(-76399026303600\), "65056873939200", \(-30721301582400\), "6173658691200"}, {"997682400", \(-35667145800\), "460929268800", \(-2996040247200\), "11185216922880", \(-25466342101200\), "35952482966400", \(-30721301582400\), "14552195486400", \(-2932487878320\)}, {\(-193993800\), "6983776800", \(-90789098400\), "593155442880", \(-2224332910800\), "5084189510400", \(-7202601806400\), "6173658691200", \(-2932487878320\), "592421793600"} }, ColumnAlignments->{Decimal}], "\[NoBreak]", ")"}], TraditionalForm]], "Output"] }, Open ]] }, Open ]], Cell[CellGroupData[{ Cell[TextData[{ "Compute and simplify ", Cell[BoxData[ \(TraditionalForm\`\[ScriptCapitalI]\_10\ .\ \[ScriptCapitalH]\_10\)]], ". Is the result as expected?" }], "NumberedExercise"], Cell[CellGroupData[{ Cell[BoxData[ \(TraditionalForm\`\[ScriptCapitalI]\_10 . \[ScriptCapitalH]\_10\)], \ "Input"], Cell[BoxData[ FormBox[ RowBox[{"(", "\[NoBreak]", GridBox[{ {"1", "0", "0", "0", "0", "0", "0", "0", "0", "0"}, {"0", "1", "0", "0", "0", "0", "0", "0", "0", "0"}, {"0", "0", "1", "0", "0", "0", "0", "0", "0", "0"}, {"0", "0", "0", "1", "0", "0", "0", "0", "0", "0"}, {"0", "0", "0", "0", "1", "0", "0", "0", "0", "0"}, {"0", "0", "0", "0", "0", "1", "0", "0", "0", "0"}, {"0", "0", "0", "0", "0", "0", "1", "0", "0", "0"}, {"0", "0", "0", "0", "0", "0", "0", "1", "0", "0"}, {"0", "0", "0", "0", "0", "0", "0", "0", "1", "0"}, {"0", "0", "0", "0", "0", "0", "0", "0", "0", "1"} }, ColumnAlignments->{Decimal}], "\[NoBreak]", ")"}], TraditionalForm]], "Output"] }, Open ]], Cell[TextData[{ "Yes. We get a ", Cell[BoxData[ \(TraditionalForm\`10\[Times]10\)]], " unit matrix." }], "NumberedAnswer"] }, Open ]], Cell[CellGroupData[{ Cell[TextData[{ "From this result, subtract off an identity matrix. 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", CounterBox["Section"], ". ", CounterBox["Subsection"], "\tMatrix Norm" }], "Subsection"], Cell[CellGroupData[{ Cell[TextData[{ "Defining the ", StyleBox["matrix norm", FontSlant->"Italic"], " as the sum of the squares of all the elements of a matrix , what answer \ will you get for (i) a zero matrix and (ii) an Identity matrix of size ", Cell[BoxData[ \(TraditionalForm\`n\[Times]n\)]], "?" }], "NumberedExercise"], Cell["\<\ (i) 0 (ii) n\ \>", "NumberedAnswer"] }, Open ]], Cell[CellGroupData[{ Cell["\<\ What can you say in general about the matrix norm of a non-zero \ matrix?\ \>", "NumberedExercise"], Cell["It will be greater than 0", "NumberedAnswer"] }, Open ]], Cell[CellGroupData[{ Cell[TextData[{ "Produce a random integer matrix of size 5\[Times]5 using ", StyleBox["Table", "Input"], " and ", StyleBox["Random", "Input"], ". Square your matrix. Apply ", StyleBox["Plus", "Input"], " to the result twice. 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Can you see how this worked? \ \>", "Text"], Cell[TextData[{ "The above operations do ", StyleBox["not", FontSlant->"Italic"], " depend on the size of the matrix! You could use ", Cell[BoxData[ \(TraditionalForm \`\[Sum]\_\(i = 1\)\%n \(\[Sum]\_\(j = 1\)\%n \[ScriptCapitalA]\_\(i, j\)\)\)]], " but then you need to compute ", Cell[BoxData[ \(TraditionalForm\`n\)]], "." }], "Warning"], Cell[CellGroupData[{ Cell[TextData[{ "Combine the above steps to create a new function for computing the matrix \ norm, ", Cell[BoxData[ \(TraditionalForm\`\[LeftDoubleBracketingBar]\[ScriptCapitalM]\ \[RightDoubleBracketingBar]\)]], ", by completing the following definition ", StyleBox["\[LeftDoubleBracketingBar]\[ScriptCapitalM]_\ \[RightDoubleBracketingBar]:=Apply[Plus,\[Ellipsis]]", "Input"], ". This function takes a single matrix as its argument. Check your \ definition on a zero matrix and a unit matrix." }], "NumberedExercise"], Cell[BoxData[ FormBox[ RowBox[{ StyleBox[\(\[LeftDoubleBracketingBar]\[ScriptCapitalM]_\ \[RightDoubleBracketingBar]\), "Input"], StyleBox[":=", "Input"], RowBox[{ StyleBox["Apply", "Input"], StyleBox["[", "Input"], RowBox[{ StyleBox["Plus", "Input"], StyleBox[",", "Input"], RowBox[{ StyleBox["Apply", "Input"], StyleBox["[", "Input"], RowBox[{ StyleBox["Plus", "Input"], StyleBox[",", "Input"], SuperscriptBox[ StyleBox["\[ScriptCapitalM]", "Input"], "2"]}], "]"}]}], StyleBox["]", "Input"]}]}], TraditionalForm]], "Input"], Cell[CellGroupData[{ Cell[BoxData[ FormBox[ StyleBox[\(\[LeftDoubleBracketingBar]Table[ 0, {5}, {5}]\[RightDoubleBracketingBar]\), "Input"], TraditionalForm]], "Input"], Cell[BoxData[ \(TraditionalForm\`0\)], "Output"] }, Open ]], Cell[CellGroupData[{ Cell[BoxData[ FormBox[ StyleBox[\(\[LeftDoubleBracketingBar]IdentityMatrix[ 5]\[RightDoubleBracketingBar]\), "Input"], TraditionalForm]], "Input"], Cell[BoxData[ \(TraditionalForm\`5\)], "Output"] }, Open ]] }, Open ]] }, Closed]], Cell[CellGroupData[{ Cell[TextData[{ CounterBox["Title"], ". 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Compute ", Cell[BoxData[ \(TraditionalForm\`\[ScriptCapitalI]\_10\ .\ \[ScriptCapitalN]\_10\)]], ". 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The diagonal elements are close to one \ but there are significant errors in the off-diagonal terms. Also we get \ warning messages when computing the inverse.\ \>", "NumberedAnswer"] }, Open ]], Cell[CellGroupData[{ Cell[TextData[{ "Realizing what ", Cell[BoxData[ \(TraditionalForm\`\[ScriptCapitalI]\_10\ . \ \ \[ScriptCapitalN]\_10\)]], " should be, subtract the exact result from your answer and apply \ \[LeftDoubleBracketingBar]_\[RightDoubleBracketingBar] to the difference. \ What does your answer tell you?" }], "NumberedExercise"], Cell[CellGroupData[{ Cell[BoxData[ \(TraditionalForm\`\[LeftDoubleBracketingBar]% - IdentityMatrix[Length[%]]\[RightDoubleBracketingBar]\)], "Input"], Cell[BoxData[ \(TraditionalForm\`8.472944132585903`*^-6\)], "Output"] }, Open ]], Cell["\<\ The norm should be zero so this value tells us how far we are away \ from the correct answer.\ \>", "NumberedAnswer"] }, Open ]], Cell[CellGroupData[{ Cell[TextData[{ "Repeat steps (10)\[Dash](12) but this time, instead of using ", StyleBox["N", "Input"], " compute the numerical values to 30 decimal places (Look up ", StyleBox["N", "Input"], " to see how to do this). Is this result expected?" }], "NumberedExercise"], Cell[BoxData[ \(TraditionalForm\`\(\[ScriptCapitalN]\_10 = N[\[ScriptCapitalH]\_10, 30];\)\)], "Input"], Cell[CellGroupData[{ Cell[BoxData[ \(TraditionalForm\`Map[Precision, %, {2}]\)], "Input"], Cell[BoxData[ FormBox[ RowBox[{"(", "\[NoBreak]", GridBox[{ {"30", "30", "30", "30", "30", "30", "30", "30", "30", "30"}, {"30", "30", "30", "30", "30", "30", "30", "30", "30", "30"}, {"30", "30", "30", "30", "30", "30", "30", "30", "30", "30"}, {"30", "30", "30", "30", "30", "30", "30", "30", "30", "30"}, {"30", "30", "30", "30", "30", "30", "30", "30", "30", "30"}, {"30", "30", "30", "30", "30", "30", "30", "30", "30", "30"}, {"30", "30", "30", "30", "30", "30", "30", "30", "30", "30"}, {"30", "30", "30", "30", "30", "30", "30", "30", "30", "30"}, {"30", "30", "30", "30", "30", "30", "30", "30", "30", "30"}, {"30", "30", "30", "30", "30", "30", "30", "30", "30", "30"} }, ColumnAlignments->{Decimal}], "\[NoBreak]", ")"}], TraditionalForm]], "Output"] }, Open ]], Cell[BoxData[ \(TraditionalForm\`\(\[ScriptCapitalI]\_10 = Inverse[\[ScriptCapitalN]\_10];\)\)], "Input"], Cell[CellGroupData[{ Cell[BoxData[ \(TraditionalForm\`\[ScriptCapitalI]\_10 . \[ScriptCapitalN]\_10\)], \ "Input"], Cell[BoxData[ FormBox[ RowBox[{"(", "\[NoBreak]", GridBox[{ {"1.000000000000000000037422`6.2716", "3.7951`-13.1011*^-20", "5.7134`-12.8804*^-20", "1.34629`-12.4691*^-19", \(-2.9003`-13.1001*^-20\), "2.6751`-13.1023*^-20", "2.46401`-12.1074*^-19", "5.0857`-12.7642*^-20", "2.42031`-12.0599*^-19", \(-7.1513`-12.5643*^-20\)}, {\(-9.435`-10.2967*^-16\), "0.9999999999999943561`4.7765", \(-6.8705`-9.3436*^-15\), \ \(-6.5065`-9.3283*^-15\), \(-4.7044`-9.4335*^-15\), \(-2.0008`-9.7719*^-15\), \ \(-3.4859`-9.5003*^-15\), \(-1.679`-9.7891*^-15\), \(-5.8357`-9.2214*^-15\), "1.6492`-9.745*^-15"}, {"6.4221`-10.5668*^-15", "1.13004`-10.2736*^-14", "1.0000000000000016448`3.7161", "2.6398`-10.8234*^-15", "5.6708`-10.4558*^-15", "4.7081`-10.5037*^-15", "1.13136`-10.0925*^-14", "1.7699`-10.8697*^-15", "1.59746`-9.8876*^-14", \(-2.6201`-10.6476*^-15\)}, {\(-3.4224`-11.6462*^-15\), \(-7.174`-12.2771*^-16\), "6.798`-11.2577*^-15", "0.9999999999999996164`2.9487", "1.9566`-11.7243*^-15", "4.8803`-11.2945*^-15", \(-4.4942`-11.2999*^-15\), "1.6304`-11.7118*^-15", \(-1.21719`-10.8121*^-14\), "6.5181`-11.0583*^-15"}, {"2.7447`-12.3084*^-15", "1.5905`-12.4978*^-15", \(-5.7851`-11.8943*^-15\), "1.9002`-12.339*^-15", "0.9999999999999997862`2.4177", \(-3.9129`-11.9571*^-15\), "4.5332`-11.8628*^-15", \(-1.5822`-12.2915*^-15\), "1.25518`-11.3654*^-14", \(-1.0118`-11.434*^-14\)}, {\(-2.7849`-12.6545*^-15\), \(-5.4806`-12.3129*^-15\), \ \(-7.4114`-12.1392*^-15\), \(-7.734`-12.082*^-15\), "9.454`-12.9593*^-16", "0.9999999999999969189`2.0978", \(-1.31707`-11.7522*^-14\), \ \(-3.3093`-12.3237*^-15\), \(-1.5037`-11.6397*^-14\), "4.3616`-12.1521*^-15"}, {"3.29`-12.7277*^-15", "5.6485`-12.4454*^-15", "1.035`-13.1398*^-15", "3.1334`-12.62*^-15", \(-2.076`-13.7634*^-16\), "1.3096`-12.9308*^-15", "1.0000000000000129686`1.9825", "3.6532`-12.4265*^-15", "1.66677`-11.7407*^-14", \(-6.3326`-12.136*^-15\)}, {\(-2.5471`-12.7668*^-15\), \(-1.7474`-12.883*^-15\), \ \(-4.082`-13.4719*^-16\), \(-5.8162`-12.2795*^-15\), "4.033`-13.4032*^-16", \(-6.336`-13.1742*^-16\), \ \(-1.18632`-11.8715*^-14\), "0.9999999999999978723`2.0826", \(-1.37992`-11.751*^-14\), "8.2114`-11.9514*^-15"}, {"2.7795`-12.4011*^-15", "7.344`-12.9318*^-16", \(-4.558`-12.0964*^-15\), "1.5082`-12.538*^-15", \(-2.5764`-12.2701*^-15\), \ \(-5.2677`-11.9268*^-15\), "5.7991`-11.8547*^-15", \(-2.758`-13.1492*^-16\), "1.000000000000014229`2.4368", \(-7.6412`-11.655*^-15\)}, {"1.2617`-12.0455*^-15", "1.1056`-12.0556*^-15", "7.4871`-11.1823*^-15", "1.2014`-11.9383*^-15", "3.8022`-11.4026*^-15", "4.7329`-11.2748*^-15", \(-1.4678`-11.753*^-15\), "6.518`-11.0772*^-15", \(-9.1671`-10.9026*^-15\), "1.0000000000000129021`3.1602"} }, ColumnAlignments->{Decimal}], "\[NoBreak]", ")"}], TraditionalForm]], "Output"] }, Open ]], Cell[CellGroupData[{ Cell[BoxData[ \(TraditionalForm\`\[LeftDoubleBracketingBar]% - IdentityMatrix[Length[%]]\[RightDoubleBracketingBar]\)], "Input"], Cell[BoxData[ \(TraditionalForm\`2.99101117311`0*^-8\)], "Output"] }, Open ]], Cell["\<\ No warning message this time. The norm is about two orders of \ magnitude smaller. \ \>", "NumberedAnswer"] }, Open ]] }, Closed]] }, Closed]], Cell[CellGroupData[{ Cell[TextData[{ CounterBox["Title"], ".", CounterBox["Section"], "\tInterpolation" }], "Section"], Cell[CellGroupData[{ Cell[TextData[{ CounterBox["Title"], ".", CounterBox["Section"], ".", CounterBox["Subsection"], "\tReading in data from an external file" }], "Subsection"], Cell[CellGroupData[{ Cell["Getting the Data", "Subsubsection"], Cell[TextData[{ "Get a copy of the file ", ButtonBox["rt.data", ButtonData:>{ URL[ "ftp://ftp.physics.uwa.edu.au/pub/Computational/rt.data"], None}, ButtonStyle->"Hyperlink"], " from the URL ", ButtonBox["ftp://ftp.physics.uwa.edu.au/pub/Computational/rt.data", ButtonData:>{ URL[ "ftp://ftp.physics.uwa.edu.au/pub/Computational/rt.data"], None}, ButtonStyle->"Hyperlink"], ". Save the resulting file as ", StyleBox["rt.data", "Input"], ". You can then use ", StyleBox["Get File Path...", "Menu"], " under the ", StyleBox["Input", "Menu"], " menu to access this file. Entering ", StyleBox["!!rt.data", "Input"], " should allow you to view this file." }], "Text"] }, Open ]], Cell[CellGroupData[{ Cell["Data Structure", "Subsubsection"], Cell[TextData[{ "You should be able to see the structure of the data file. (", StyleBox["Hint:", FontSlant->"Italic"], " the data was taken from mounting an object on a turn-table and taking \ measurements of the radial length (", Cell[BoxData[ \(TraditionalForm\`r\)]], ") every 5 degrees (", Cell[BoxData[ \(TraditionalForm\`\[Theta]\)]], ")). " }], "Text"], Cell[CellGroupData[{ Cell[TextData[{ "Look up ", StyleBox["ReadList", "Input"], " and use it to read in ", StyleBox["rt.data", "Input"], " in an appropriate format. Save the result as r\[Theta]. ", StyleBox["Hint", FontSlant->"Italic"], ": you will want to read in the data as ", StyleBox["pairs ", FontSlant->"Italic"], "of numbers. You might want to look at the on-line documentation for ", StyleBox["ReadList", "Input"], "." }], "NumberedExercise"], Cell[CellGroupData[{ Cell[BoxData[ \(TraditionalForm\`r\[Theta] = ReadList["\", {Number, Number}]\)], "Input"], Cell[BoxData[ FormBox[ RowBox[{"(", "\[NoBreak]", GridBox[{ {"2.1006`", "0"}, {"2.104636986431339`", "5"}, {"2.108252019049731`", "10"}, {"2.112901939089985`", "15"}, {"2.120131343946979`", "20"}, {"2.13181906563078`", "25"}, {"2.149656616218929`", "30"}, {"2.175131353695856`", "35"}, {"2.20919762960846`", "40"}, {"2.251929076439512`", "45"}, {"2.301996249185268`", "50"}, {"2.354935292007386`", "55"}, {"2.400647236491379`", "60"}, {"2.423719537805815`", "65"}, {"2.411515123894715`", "70"}, {"2.365884212465177`", "75"}, {"2.301589541981545`", "80"}, {"2.233466320947635`", "85"}, {"2.170221257712685`", "90"}, {"2.117181599248319`", "95"}, {"2.075468526668255`", "100"}, {"2.046100791561403`", "105"}, {"2.029027849058653`", "110"}, {"2.0242211736236`", "115"}, {"2.032030573857567`", "120"}, {"2.05248488077628`", "125"}, {"2.086193785593291`", "130"}, {"2.133939811044736`", "135"}, {"2.196335359863972`", "140"}, {"2.274383188873538`", "145"}, {"2.368226863410824`", "150"}, {"2.477563435372177`", "155"}, {"2.59810109800822`", "160"}, {"2.720760339919281`", "165"}, {"2.827725142216809`", "170"}, {"2.891811324418465`", "175"}, {"2.889545827964561`", "180"}, {"2.819557197061996`", "185"}, {"2.704273284378449`", "190"}, {"2.572078414597517`", "195"}, {"2.442185540571726`", "200"}, {"2.324281521431907`", "205"}, {"2.222847059058453`", "210"}, {"2.138004761945429`", "215"}, {"2.069408043178231`", "220"}, {"2.015800375952927`", "225"}, {"1.976364493527556`", "230"}, {"1.950155986858029`", "235"}, {"1.936364871177665`", "240"}, {"1.93483876764266`", "245"}, {"1.945179987246012`", "250"}, {"1.967274106083634`", "255"}, {"2.001368556046825`", "260"}, {"2.046700397643405`", "265"}, {"2.103000757550927`", "270"}, {"2.167187236546002`", "275"}, {"2.233573777929696`", "280"}, {"2.291832669702183`", "285"}, {"2.327137221159409`", "290"}, {"2.329463702991894`", "295"}, {"2.302728944466888`", "300"}, {"2.260284595973223`", "305"}, {"2.214746936574824`", "310"}, {"2.17363955527591`", "315"}, {"2.140094818281815`", "320"}, {"2.115038283291072`", "325"}, {"2.098304328536158`", "330"}, {"2.089000201343451`", "335"}, {"2.085755529076004`", "340"}, {"2.086988525387166`", "345"}, {"2.090885974617337`", "350"}, {"2.095920608122359`", "355"}, {"2.1006`", "360"} }, ColumnAlignments->{Decimal}], "\[NoBreak]", ")"}], TraditionalForm]], "Output"] }, Closed]] }, Closed]], Cell[TextData[{ "Note that the valid format specifications are ", StyleBox["Byte", "Input"], ", ", StyleBox["Character", "Input"], ", ", StyleBox["Expression", "Input"], ", ", StyleBox["Number", "Input"], ", ", StyleBox["Real", "Input"], ", ", StyleBox["Record", "Input"], ", ", StyleBox["String", "Input"], ", and ", StyleBox["Word", "Input"], "." }], "Warning"], Cell[CellGroupData[{ Cell[TextData[{ "Enter ", StyleBox["{rs,\[Theta]s}=Transpose[r\[Theta]]", "Input"], ". What does ", StyleBox["Transpose", "Input"], " do? 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The output shows the range of \[Theta] values over which the \ interpolation has been defined. Use ", StyleBox["\[ScriptR][0.2]", "Input"], " to compute the (interpolated) radius at an angle of 0.2 radians. 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The integral over ", StyleBox["r", FontSlant->"Italic"], " is trivial:" }], "Text"], Cell[CellGroupData[{ Cell[TextData[{ "Compute the inner integral ", Cell[BoxData[ \(TraditionalForm\`\[Integral]\_0\%\(f[\[Theta]]\)r \ \[DifferentialD]r\)]] }], "NumberedExercise"], Cell[CellGroupData[{ Cell[BoxData[ \(TraditionalForm\`\[Integral]\_0\%\(f[\[Theta]]\)r \[DifferentialD]r\)], \ "Input"], Cell[BoxData[ \(TraditionalForm\`\(f(\[Theta])\)\^2\/2\)], "Output"] }, Open ]] }, Closed]], Cell[CellGroupData[{ Cell[TextData[{ "From this result, the definition of the (numerical) area of a shape \ parametrized by ", Cell[BoxData[ \(TraditionalForm\`f\)]], " is " }], "Text"], Cell[BoxData[ \(TraditionalForm\`\[ScriptCapitalA](\[ScriptF]_) := 1\/2\ NIntegrate[\(\[ScriptF](\[Theta])\)\^2, {\[Theta], 0, 2\ \[Pi]}]\)], "Input"] }, Open ]], Cell[CellGroupData[{ Cell["\<\ Compute the area of the shape above by entering \[ScriptCapitalA][\ \[ScriptR]]. By looking at the graphical output above, explain whether your \ answer is reasonable?\ \>", "NumberedExercise"], Cell[CellGroupData[{ Cell[BoxData[ \(TraditionalForm\`\[ScriptCapitalA][\[ScriptR]]\)], "Input"], Cell[BoxData[ \(TraditionalForm\`15.782858528246233`\)], "Output"] }, Open ]], Cell["\<\ Counting the boxes in the plot shows that this answer is reasonable\ \ \>", "NumberedAnswer"] }, Closed]], Cell[TextData[{ "Of course, this is only an ", StyleBox["approximation", FontSlant->"Italic"], " to the area of the object as the dataset was incomplete and interpolation \ has been used." }], "Warning"] }, Closed]], Cell[CellGroupData[{ Cell[TextData[{ CounterBox["Title"], ".", CounterBox["Section"], ".", CounterBox["Subsection"], "\tCentre of mass" }], "Subsection"], Cell[CellGroupData[{ Cell[TextData[{ "The ", Cell[BoxData[ \(TraditionalForm\`x\)]], "-coordinate of the ", StyleBox["centre of mass", FontSlant->"Italic"], " of an object whose radius ", Cell[BoxData[ \(TraditionalForm\`r\)]], " is expressed as a function of ", Cell[BoxData[ \(TraditionalForm\`\[Theta]\)]], ", is given by the following integral average:" }], "Text"], Cell[BoxData[ \(TraditionalForm\`x\_CM = \(\(\[Integral]\_0\%\(2 \ \[Pi]\)\(\[Integral]\_0\%\(f[\[Theta]]\)x\ r\ \[DifferentialD]r\ \ \[DifferentialD]\[Theta]\)\)\/\(\[Integral]\_0\%\(2 \ \[Pi]\)\(\[Integral]\_0\%\(f[\[Theta]]\)\ r\ \[DifferentialD]r\ \ \[DifferentialD]\[Theta]\)\) = \(\[Integral]\_0\%\(2 \[Pi]\)\(\[Integral]\_0\ \%\(f[\[Theta]]\)\ \(r\^2\) \(cos(\[Theta])\)\ \[DifferentialD]r\ \ \[DifferentialD]\[Theta]\)\)\/\[ScriptCapitalA]\)\)], "NumberedEquation"] }, Open ]], Cell[CellGroupData[{ Cell[TextData[{ "What is the expression for the ", Cell[BoxData[ \(TraditionalForm\`y\)]], "-coordinate of the centre of mass?" }], "NumberedExercise"], Cell[TextData[Cell[BoxData[ \(TraditionalForm\`y\_CM = \(\(\[Integral]\_0\%\(2 \ \[Pi]\)\(\[Integral]\_0\%\(f[\[Theta]]\)y\ r\ \[DifferentialD]r\ \ \[DifferentialD]\[Theta]\)\)\/\(\[Integral]\_0\%\(2 \ \[Pi]\)\(\[Integral]\_0\%\(f[\[Theta]]\)\ r\ \[DifferentialD]r\ \ \[DifferentialD]\[Theta]\)\) = \(\[Integral]\_0\%\(2 \ \[Pi]\)\((\[Integral]\_0\%\(f[\[Theta]]\)r\^2\ \[DifferentialD]r\ )\) \(sin(\ \[Theta])\) \[DifferentialD]\[Theta]\)\/\[ScriptCapitalA]\)\)]]], \ "NumberedAnswer"] }, Closed]], Cell[CellGroupData[{ Cell[TextData[{ "Noting that ", Cell[BoxData[ \(TraditionalForm\`x = r\ \(cos(\[Theta])\)\)]], " and ", Cell[BoxData[ \(TraditionalForm\`y = r\ \(sin(\[Theta])\)\)]], ", modify the steps in \[Section]", CounterBox["Title"], ". ", CounterBox["Section"], ". 3 to define functions ", Cell[BoxData[ \(TraditionalForm\`\[ScriptX]\_CM[\[ScriptF]_]\)]], " and ", Cell[BoxData[ \(TraditionalForm\`\[ScriptY]\_CM[\[ScriptF]_]\)]], ". Compute the ", Cell[BoxData[ \(TraditionalForm\`x\)]], " and ", Cell[BoxData[ \(TraditionalForm\`y\)]], " coordinates of the centre of mass. By looking at the graphical output \ above, is your answer reasonable? (Approximately where should the centre of \ mass of the shape be located?)" }], "NumberedExercise"], Cell[CellGroupData[{ Cell[BoxData[ \(TraditionalForm\`\[Integral]\_0\%\(f[\[Theta]]\)\(r\^2\) \ \[DifferentialD]r\)], "Input"], Cell[BoxData[ \(TraditionalForm\`\(f(\[Theta])\)\^3\/3\)], "Output"] }, Closed]], Cell[TextData[{ "From this result, the definition of the (numerical) area of a shape \ parametrized by ", Cell[BoxData[ \(TraditionalForm\`f\)]], " is " }], "NumberedAnswer"], Cell[BoxData[ \(TraditionalForm\`\(x\_CM\)(\[ScriptF]_) := \ NIntegrate[\(\(\[ScriptF](\[Theta])\)\^3\) \(cos(\[Theta])\), \ {\[Theta], 0, 2\ \[Pi]}]\/\(3 \[ScriptCapitalA][\[ScriptF]]\)\)], "Input"], Cell[BoxData[ \(TraditionalForm\`\(y\_CM\)(\[ScriptF]_) := NIntegrate[\(\(\[ScriptF](\[Theta])\)\^3\) \(sin(\[Theta])\), \ {\[Theta], 0, 2\ \[Pi]}]\/\(3 \[ScriptCapitalA][\[ScriptF]]\)\)], "Input"], Cell[CellGroupData[{ Cell[BoxData[ \(TraditionalForm\`\(x\_CM\)(\[ScriptR])\)], "Input"], Cell[BoxData[ FormBox[ RowBox[{\(NIntegrate::"ncvb"\), \(\(:\)\(\ \)\), "\<\"NIntegrate failed \ to converge to prescribed accuracy after \\!\\(TraditionalForm\\`7\\) \ recursive bisections in \\!\\(TraditionalForm\\`\[Theta]\\) near \ \\!\\(TraditionalForm\\`\[Theta]\\) = \ \\!\\(TraditionalForm\\`3.067961575771282`\\).\"\>"}], TraditionalForm]], "Message"], Cell[BoxData[ \(TraditionalForm\`\(-0.14941977545011206`\)\)], "Output"] }, Open ]], Cell[CellGroupData[{ Cell[BoxData[ \(TraditionalForm\`\(y\_CM\)(\[ScriptR])\)], "Input"], Cell[BoxData[ FormBox[ RowBox[{\(NIntegrate::"ncvb"\), \(\(:\)\(\ \)\), "\<\"NIntegrate failed \ to converge to prescribed accuracy after \\!\\(TraditionalForm\\`7\\) \ recursive bisections in \\!\\(TraditionalForm\\`\[Theta]\\) near \ \\!\\(TraditionalForm\\`\[Theta]\\) = \ \\!\\(TraditionalForm\\`1.1535535524900022`\\).\"\>"}], TraditionalForm]], "Message"], Cell[BoxData[ \(TraditionalForm\`0.05356896279573823`\)], "Output"] }, Open ]], Cell["\<\ From the graph, these values appear reasonable. The warning \ messages arise because of the accuracy goal. If we reduce this we get a \ similar answer but with no warning message:\ \>", "NumberedAnswer"], Cell[CellGroupData[{ Cell[BoxData[ \(TraditionalForm\`Options[NIntegrate]\)], "Input"], Cell[BoxData[ FormBox[ RowBox[{"{", RowBox[{ RowBox[{"AccuracyGoal", "\[Rule]", InterpretationBox["\[Infinity]", DirectedInfinity[ 1]]}], ",", \(Compiled \[Rule] True\), ",", \(GaussPoints \[Rule] Automatic\), ",", \(MaxPoints \[Rule] Automatic\), ",", \(MaxRecursion \[Rule] 6\), ",", \(Method \[Rule] Automatic\), ",", \(MinRecursion \[Rule] 0\), ",", \(PrecisionGoal \[Rule] Automatic\), ",", \(SingularityDepth \[Rule] 4\), ",", \(WorkingPrecision \[Rule] 16\)}], "}"}], TraditionalForm]], "Output"] }, Open ]], Cell[CellGroupData[{ Cell[BoxData[ \(TraditionalForm\`NIntegrate[\(\(\[ScriptR](\[Theta])\)\^3\) \(cos(\ \[Theta])\), {\[Theta], 0, 2\ \[Pi]}, AccuracyGoal \[Rule] 3]\/\(3 \ \[ScriptCapitalA][\[ScriptR]]\)\)], "Input"], Cell[BoxData[ \(TraditionalForm\`\(-0.14941826668201763`\)\)], "Output"] }, Open ]] }, Closed]] }, Closed]] }, Open ]], Cell[CellGroupData[{ Cell[TextData[{ CounterBox["Title"], ".", CounterBox["Section"], "\tNumerical Differentiation" }], "Section"], Cell[CellGroupData[{ Cell[TextData[{ CounterBox["Title"], ". ", CounterBox["Section"], ". ", CounterBox["Subsection"], "\tNumerical derivatives" }], "Subsection"], Cell[TextData[{ "To solve differential equations ", StyleBox["numerically", FontSlant->"Italic"], ", a method for computing derivatives numerically is required. How can the \ derivative of a function be computed numerically? " }], "Text"], Cell[CellGroupData[{ Cell[TextData[{ "Compute the Taylor series of ", Cell[BoxData[ \(TraditionalForm\`\((f[x + h]\ - \ f[x])\)/h\)]], " using ", Cell[BoxData[ \(\(f[x + h] - f[x]\)\/h + O[h]\^3\)]], ". What happens to this expression when ", StyleBox["h\[Rule]0", "Input"], "?" }], "NumberedExercise"], Cell[CellGroupData[{ Cell[BoxData[ \(TraditionalForm\`\(f(h + x) - f(x)\)\/h + \(O(h)\)\^3\)], "Input"], Cell[BoxData[ FormBox[ InterpretationBox[ RowBox[{ RowBox[{ SuperscriptBox["f", "\[Prime]", MultilineFunction->None], "(", "x", ")"}], "+", RowBox[{\(1\/2\), " ", RowBox[{ SuperscriptBox["f", "\[Prime]\[Prime]", MultilineFunction->None], "(", "x", ")"}], " ", "h"}], "+", RowBox[{\(1\/6\), " ", RowBox[{ SuperscriptBox["f", TagBox[\((3)\), Derivative], MultilineFunction->None], "(", "x", ")"}], " ", \(h\^2\)}], "+", InterpretationBox[\(O(h\^3)\), SeriesData[ h, 0, {}, 0, 3, 1]]}], SeriesData[ h, 0, { Derivative[ 1][ f][ x], Times[ Rational[ 1, 2], Derivative[ 2][ f][ x]], Times[ Rational[ 1, 6], Derivative[ 3][ f][ x]]}, 0, 3, 1]], TraditionalForm]], "Output"] }, Open ]], Cell[TextData[{ "As ", Cell[BoxData[ \(TraditionalForm\`h \[Rule] 0\)]], " this tends to ", Cell[BoxData[ \(TraditionalForm\`\(f\^\[Prime]\)(x)\)]], " and the error term is proportional to ", Cell[BoxData[ \(TraditionalForm\`h\)]], "." }], "NumberedAnswer"] }, Open ]], Cell[TextData[{ "This result, with modification is the basis of techniques for integrating \ differential equations. For computing the derivative of a function ", Cell[BoxData[ \(TraditionalForm\`f\)]], " numerically, relying on this definition of differentiation is not \ optimal. As ", Cell[BoxData[ \(TraditionalForm\`h \[Rule] 0\)]], ", the error term in the Taylor series is ", StyleBox["first order", FontSlant->"Italic"], " in ", Cell[BoxData[ \(TraditionalForm\`h\)]], ". " }], "Text"], Cell[CellGroupData[{ Cell[TextData[{ "Compute the Taylor series of ", Cell[BoxData[ \(TraditionalForm\`\((f[x + h]\ - \ f[x - h])\)/\((2 h)\)\)]], " and explain why it is preferable for numerical computation of the \ derivative." }], "NumberedExercise"], Cell[CellGroupData[{ Cell[BoxData[ \(TraditionalForm\`\(f(h + x) - f(x - h)\)\/\(2\ h\) + \(O(h)\)\^3\)], \ "Input"], Cell[BoxData[ FormBox[ InterpretationBox[ RowBox[{ RowBox[{ SuperscriptBox["f", "\[Prime]", MultilineFunction->None], "(", "x", ")"}], "+", RowBox[{\(1\/6\), " ", RowBox[{ SuperscriptBox["f", TagBox[\((3)\), Derivative], MultilineFunction->None], "(", "x", ")"}], " ", \(h\^2\)}], "+", InterpretationBox[\(O(h\^3)\), SeriesData[ h, 0, {}, 0, 3, 1]]}], SeriesData[ h, 0, { Derivative[ 1][ f][ x], 0, Times[ Rational[ 1, 6], Derivative[ 3][ f][ x]]}, 0, 3, 1]], TraditionalForm]], "Output"] }, Open ]], Cell[TextData[{ "The error is now second order in ", Cell[BoxData[ \(TraditionalForm\`h\)]], "." }], "NumberedAnswer"] }, Open ]], Cell[CellGroupData[{ Cell[TextData[{ "Suppose that you can easily compute (or know) the values ", Cell[BoxData[ \(TraditionalForm\`f[x]\)]], ", ", Cell[BoxData[ \(TraditionalForm\`f[x + h]\)]], " and ", Cell[BoxData[ \(TraditionalForm\`f[x - h]\)]], ". Form an arbitrary linear combination of these terms, say " }], "Text"], Cell[BoxData[ \(TraditionalForm\`\(\[ScriptL] = \(\[Beta]\ \(f(x - h)\) + \[Gamma]\ \ \(f(h + x)\) + \[Alpha]\ \(f(x)\)\)\/h;\)\)], "Input"] }, Open ]], Cell[CellGroupData[{ Cell[TextData[{ "Expand \[ScriptL] in a Taylor series about ", Cell[BoxData[ \(TraditionalForm\`h = 0\)]], " to second order by adding ", Cell[BoxData[ \(O[h]\^2\)]], " to \[ScriptL]. Equate this (using ==) with ", Cell[BoxData[ \(TraditionalForm\`\(f\^\[Prime]\)[x]\)]], ". Use ", StyleBox["Solve", "Input"], " to find \[Alpha], \[Beta], and \[Gamma] and thus determine the best \ numerical expression for computing ", Cell[BoxData[ \(TraditionalForm\`\(f\^\[Prime]\)[x]\)]], "from ", Cell[BoxData[ \(TraditionalForm\`f[x]\)]], ", f[", Cell[BoxData[ \(TraditionalForm\`\(\(\(x + h\) ]\)\(\ \)\)\)]], "and ", Cell[BoxData[ \(TraditionalForm\`f[x - h]\)]], ". Substitute your values back into \[ScriptL]." }], "NumberedExercise"], Cell[CellGroupData[{ Cell[BoxData[ \(TraditionalForm\`\[ScriptL] + \(O(h)\)\^2\)], "Input"], Cell[BoxData[ FormBox[ InterpretationBox[ RowBox[{\(\(\[Alpha]\ \(f(x)\) + \[Beta]\ \(f(x)\) + \[Gamma]\ \(f( x)\)\)\/h\), "+", RowBox[{"(", RowBox[{ RowBox[{"\[Gamma]", " ", RowBox[{ SuperscriptBox["f", "\[Prime]", MultilineFunction->None], "(", "x", ")"}]}], "-", RowBox[{"\[Beta]", " ", RowBox[{ SuperscriptBox["f", "\[Prime]", MultilineFunction->None], "(", "x", ")"}]}]}], ")"}], "+", RowBox[{ RowBox[{"(", RowBox[{ RowBox[{\(1\/2\), " ", "\[Beta]", " ", RowBox[{ SuperscriptBox["f", "\[Prime]\[Prime]", MultilineFunction->None], "(", "x", ")"}]}], "+", RowBox[{\(1\/2\), " ", "\[Gamma]", " ", RowBox[{ SuperscriptBox["f", "\[Prime]\[Prime]", MultilineFunction->None], "(", "x", ")"}]}]}], ")"}], " ", "h"}], "+", InterpretationBox[\(O(h\^2)\), SeriesData[ h, 0, {}, -1, 2, 1]]}], SeriesData[ h, 0, { Plus[ Times[ \[Alpha], f[ x]], Times[ \[Beta], f[ x]], Times[ \[Gamma], f[ x]]], Plus[ Times[ -1, \[Beta], Derivative[ 1][ f][ x]], Times[ \[Gamma], Derivative[ 1][ f][ x]]], Plus[ Times[ Rational[ 1, 2], \[Beta], Derivative[ 2][ f][ x]], Times[ Rational[ 1, 2], \[Gamma], Derivative[ 2][ f][ x]]]}, -1, 2, 1]], TraditionalForm]], "Output"] }, Open ]], Cell[CellGroupData[{ Cell[BoxData[ \(TraditionalForm\`Solve[% \[Equal] \(f\^\[Prime]\)( x), {\[Alpha], \[Beta], \[Gamma]}]\)], "Input"], Cell[BoxData[ \(TraditionalForm\`{{\[Alpha] \[Rule] 0, \[Beta] \[Rule] \(-\(1\/2\)\), \[Gamma] \[Rule] 1\/2}}\)], "Output"] }, Open ]], Cell[CellGroupData[{ Cell[BoxData[ \(TraditionalForm\`\[ScriptL] /. First[%] // Simplify\)], "Input"], Cell[BoxData[ \(TraditionalForm\`\(f(h + x) - f(x - h)\)\/\(2\ h\)\)], "Output"] }, Open ]] }, Open ]], Cell[TextData[{ "You should have found the best \"two-point\" approximation to ", Cell[BoxData[ \(TraditionalForm\`\(f\^\[Prime]\)[x]\)]], " which is ", Cell[BoxData[ \(TraditionalForm\`\(\ f[h + x] - \ f[x - h]\)\/\(2 h\)\)]], "." }], "Text"], Cell[CellGroupData[{ Cell[TextData[{ "Expand the best \"two-point\" approximation to ", Cell[BoxData[ \(TraditionalForm\`\(f\^\[Prime]\)[x]\)]], " in a Taylor series about ", Cell[BoxData[ \(TraditionalForm\`h = 0\)]], " to third order by adding ", Cell[BoxData[ \(O[h]\^3\)]], "to it. What is the size of the error term? Why is this expression better \ than the simple ", Cell[BoxData[ \(TraditionalForm\`\((f[x + h] - f[x])\)/h\)]], " for approximately computing the derivative?" }], "NumberedExercise"], Cell[CellGroupData[{ Cell[BoxData[ \(TraditionalForm\`\(f(h + x) - f(x - h)\)\/\(2\ h\) + \(O(h)\)\^3\)], \ "Input"], Cell[BoxData[ FormBox[ InterpretationBox[ RowBox[{ RowBox[{ SuperscriptBox["f", "\[Prime]", MultilineFunction->None], "(", "x", ")"}], "+", RowBox[{\(1\/6\), " ", RowBox[{ SuperscriptBox["f", TagBox[\((3)\), Derivative], MultilineFunction->None], "(", "x", ")"}], " ", \(h\^2\)}], "+", InterpretationBox[\(O(h\^3)\), SeriesData[ h, 0, {}, 0, 3, 1]]}], SeriesData[ h, 0, { Derivative[ 1][ f][ x], 0, Times[ Rational[ 1, 6], Derivative[ 3][ f][ x]]}, 0, 3, 1]], TraditionalForm]], "Output"] }, Open ]], Cell[TextData[{ "The error term is ", Cell[BoxData[ FormBox[ RowBox[{\(1\/6\), " ", RowBox[{ SuperscriptBox["f", TagBox[\((3)\), Derivative], MultilineFunction->None], "(", "x", ")"}], " ", \(h\^2\)}], TraditionalForm]]], " which is second order in ", Cell[BoxData[ \(TraditionalForm\`h\)]], "." }], "NumberedAnswer"] }, Open ]], Cell[CellGroupData[{ Cell[TextData[{ "For ", Cell[BoxData[ \(TraditionalForm\`sin(x)\)]], " compute and compare: ", Cell[BoxData[ FormBox[ RowBox[{ SuperscriptBox["Sin", "\[Prime]", MultilineFunction->None], "[", "x", "]"}], TraditionalForm]]], ", ", Cell[BoxData[ \(TraditionalForm\`\((Sin[x + h] - Sin[x])\)/h\)]], " and your best expression above for x\[Rule]0.2, h\[Rule]0.1." }], "NumberedExercise"], Cell[CellGroupData[{ Cell[BoxData[ \(TraditionalForm\`exact = \(sin\^\[Prime]\)(x) /. x \[Rule] 0.2\)], "Input"], Cell[BoxData[ \(TraditionalForm\`0.9800665778412416`\)], "Output"] }, Open ]], Cell[CellGroupData[{ Cell[BoxData[ \(TraditionalForm\`unsymmetric = \(sin(h + x) - sin(x)\)\/h /. {x \[Rule] 0.2, h \[Rule] 0.1}\)], "Input"], Cell[BoxData[ \(TraditionalForm\`0.9685087586627836`\)], "Output"] }, Open ]], Cell[CellGroupData[{ Cell[BoxData[ \(TraditionalForm\`twopoint = \(sin(h + x) - sin(x - h)\)\/\(2\ h\) /. {x \ \[Rule] 0.2, h \[Rule] 0.1}\)], "Input"], Cell[BoxData[ \(TraditionalForm\`0.9784339500725572`\)], "Output"] }, Open ]], Cell[CellGroupData[{ Cell[BoxData[ \(TraditionalForm\`exact - unsymmetric\)], "Input"], Cell[BoxData[ \(TraditionalForm\`0.011557819178457995`\)], "Output"] }, Open ]], Cell[CellGroupData[{ Cell[BoxData[ \(TraditionalForm\`exact - twopoint\)], "Input"], Cell[BoxData[ \(TraditionalForm\`0.001632627768684447`\)], "Output"] }, Open ]], Cell["\<\ It is clear that the two point approximation is much better than \ the unsymmetric one.\ \>", "NumberedAnswer"] }, Open ]], Cell[CellGroupData[{ Cell[TextData[{ "Obtain the best expression for computing ", Cell[BoxData[ \(TraditionalForm\`\(f\^\[Prime]\)[x]\)]], " numerically in terms of ", Cell[BoxData[ \(TraditionalForm\`f[x + 2 h]\)]], ", ", Cell[BoxData[ \(TraditionalForm\`f[x + h]\)]], ", ", Cell[BoxData[ \(TraditionalForm\`f[x]\)]], ", ", Cell[BoxData[ \(TraditionalForm\`f[x - h]\)]], ", and ", Cell[BoxData[ \(TraditionalForm\`f[x - 2 h]\)]], ". Also determine the error term." }], "NumberedExercise"], Cell[CellGroupData[{ Cell[BoxData[ \(TraditionalForm\`\((\[ScriptL] = \(\[Alpha]\ \(f(x)\) + \[Beta]\ \(f(x \ - h)\) + \[Gamma]\ \(f(x + h)\) + \[Delta]\ \(f(x - 2 h)\) + \[Epsilon]\ \ \(f(x + 2 h)\)\)\/h)\) + \(O(h)\)\^4\)], "Input"], Cell[BoxData[ FormBox[ InterpretationBox[ RowBox[{\(\(\[Alpha]\ \(f(x)\) + \[Beta]\ \(f(x)\) + \[Gamma]\ \(f( x)\) + \[Delta]\ \(f(x)\) + \[Epsilon]\ \(f(x)\)\)\/h\), "+", RowBox[{"(", RowBox[{ RowBox[{\(-\[Beta]\), " ", RowBox[{ SuperscriptBox["f", "\[Prime]", MultilineFunction->None], "(", "x", ")"}]}], "+", RowBox[{"\[Gamma]", " ", RowBox[{ SuperscriptBox["f", "\[Prime]", MultilineFunction->None], "(", "x", ")"}]}], "-", RowBox[{"2", " ", "\[Delta]", " ", RowBox[{ SuperscriptBox["f", "\[Prime]", MultilineFunction->None], "(", "x", ")"}]}], "+", RowBox[{"2", " ", "\[Epsilon]", " ", RowBox[{ SuperscriptBox["f", "\[Prime]", MultilineFunction->None], "(", "x", ")"}]}]}], ")"}], "+", RowBox[{ RowBox[{"(", RowBox[{ RowBox[{\(1\/2\), " ", "\[Beta]", " ", RowBox[{ SuperscriptBox["f", "\[Prime]\[Prime]", MultilineFunction->None], "(", "x", ")"}]}], "+", RowBox[{\(1\/2\), " ", "\[Gamma]", " ", RowBox[{ SuperscriptBox["f", "\[Prime]\[Prime]", MultilineFunction->None], "(", "x", ")"}]}], "+", RowBox[{"2", " ", "\[Delta]", " ", RowBox[{ SuperscriptBox["f", "\[Prime]\[Prime]", MultilineFunction->None], "(", "x", ")"}]}], "+", RowBox[{"2", " ", "\[Epsilon]", " ", RowBox[{ SuperscriptBox["f", "\[Prime]\[Prime]", MultilineFunction->None], "(", "x", ")"}]}]}], ")"}], " ", "h"}], "+", RowBox[{ RowBox[{"(", RowBox[{ RowBox[{\(-\(1\/6\)\), " ", "\[Beta]", " ", RowBox[{ SuperscriptBox["f", TagBox[\((3)\), Derivative], MultilineFunction->None], "(", "x", ")"}]}], "+", RowBox[{\(1\/6\), " ", "\[Gamma]", " ", RowBox[{ SuperscriptBox["f", TagBox[\((3)\), Derivative], MultilineFunction->None], "(", "x", ")"}]}], "-", RowBox[{\(4\/3\), " ", "\[Delta]", " ", RowBox[{ SuperscriptBox["f", TagBox[\((3)\), Derivative], MultilineFunction->None], "(", "x", ")"}]}], "+", RowBox[{\(4\/3\), " ", "\[Epsilon]", " ", RowBox[{ SuperscriptBox["f", TagBox[\((3)\), Derivative], MultilineFunction->None], "(", "x", ")"}]}]}], ")"}], " ", \(h\^2\)}], "+", RowBox[{ RowBox[{"(", RowBox[{ RowBox[{\(1\/24\), " ", "\[Beta]", " ", RowBox[{ SuperscriptBox["f", TagBox[\((4)\), Derivative], MultilineFunction->None], "(", "x", ")"}]}], "+", RowBox[{\(1\/24\), " ", "\[Gamma]", " ", RowBox[{ SuperscriptBox["f", TagBox[\((4)\), Derivative], MultilineFunction->None], "(", "x", ")"}]}], "+", RowBox[{\(2\/3\), " ", "\[Delta]", " ", RowBox[{ SuperscriptBox["f", TagBox[\((4)\), Derivative], MultilineFunction->None], "(", "x", ")"}]}], "+", RowBox[{\(2\/3\), " ", "\[Epsilon]", " ", RowBox[{ SuperscriptBox["f", TagBox[\((4)\), Derivative], MultilineFunction->None], "(", "x", ")"}]}]}], ")"}], " ", \(h\^3\)}], "+", InterpretationBox[\(O(h\^4)\), SeriesData[ h, 0, {}, -1, 4, 1]]}], SeriesData[ h, 0, { Plus[ Times[ \[Alpha], f[ x]], Times[ \[Beta], f[ x]], Times[ \[Gamma], f[ x]], Times[ \[Delta], f[ x]], Times[ \[Epsilon], f[ x]]], Plus[ Times[ -1, \[Beta], Derivative[ 1][ f][ x]], Times[ \[Gamma], Derivative[ 1][ f][ x]], Times[ -2, \[Delta], Derivative[ 1][ f][ x]], Times[ 2, \[Epsilon], Derivative[ 1][ f][ x]]], Plus[ Times[ Rational[ 1, 2], \[Beta], Derivative[ 2][ f][ x]], Times[ Rational[ 1, 2], \[Gamma], Derivative[ 2][ f][ x]], Times[ 2, \[Delta], Derivative[ 2][ f][ x]], Times[ 2, \[Epsilon], Derivative[ 2][ f][ x]]], Plus[ Times[ Rational[ -1, 6], \[Beta], Derivative[ 3][ f][ x]], Times[ Rational[ 1, 6], \[Gamma], Derivative[ 3][ f][ x]], Times[ Rational[ -4, 3], \[Delta], Derivative[ 3][ f][ x]], Times[ Rational[ 4, 3], \[Epsilon], Derivative[ 3][ f][ x]]], Plus[ Times[ Rational[ 1, 24], \[Beta], Derivative[ 4][ f][ x]], Times[ Rational[ 1, 24], \[Gamma], Derivative[ 4][ f][ x]], Times[ Rational[ 2, 3], \[Delta], Derivative[ 4][ f][ x]], Times[ Rational[ 2, 3], \[Epsilon], Derivative[ 4][ f][ x]]]}, -1, 4, 1]], TraditionalForm]], "Output"] }, Open ]], Cell[CellGroupData[{ Cell[BoxData[ \(TraditionalForm\`Solve[% \[Equal] \(f\^\[Prime]\)( x), {\[Alpha], \[Beta], \[Gamma], \[Delta], \[Epsilon]}]\)], "Input"], Cell[BoxData[ \(TraditionalForm\`{{\[Alpha] \[Rule] 0, \[Beta] \[Rule] \(-\(2\/3\)\), \[Gamma] \[Rule] 2\/3, \[Delta] \[Rule] 1\/12, \[Epsilon] \[Rule] \(-\(1\/12\)\)}}\)], "Output"] }, Open ]], Cell[CellGroupData[{ Cell[BoxData[ \(TraditionalForm\`\[ScriptL] /. First[%] // Simplify\)], "Input"], Cell[BoxData[ \(TraditionalForm\`\(f(x - 2\ h) - 8\ \(f(x - h)\) + 8\ \(f(h + x)\) - \ f(2\ h + x)\)\/\(12\ h\)\)], "Output"] }, Open ]], Cell["Expanding this into a series we get the error term:", "NumberedAnswer"], Cell[CellGroupData[{ Cell[BoxData[ \(TraditionalForm\`% + \(O(h)\)\^5\)], "Input"], Cell[BoxData[ FormBox[ InterpretationBox[ RowBox[{ RowBox[{ SuperscriptBox["f", "\[Prime]", MultilineFunction->None], "(", "x", ")"}], "-", RowBox[{\(1\/30\), " ", RowBox[{ SuperscriptBox["f", TagBox[\((5)\), Derivative], MultilineFunction->None], "(", "x", ")"}], " ", \(h\^4\)}], "+", InterpretationBox[\(O(h\^5)\), SeriesData[ h, 0, {}, 0, 5, 1]]}], SeriesData[ h, 0, { Derivative[ 1][ f][ x], 0, 0, 0, Times[ Rational[ -1, 30], Derivative[ 5][ f][ x]]}, 0, 5, 1]], TraditionalForm]], "Output"] }, Open ]] }, Open ]], Cell[CellGroupData[{ Cell[TextData[{ "Obtain the best expression for computing ", Cell[BoxData[ \(TraditionalForm\`\(f\^\[DoublePrime]\)[x]\)]], " numerically in terms of ", Cell[BoxData[ \(TraditionalForm\`f[x + h]\)]], ", ", Cell[BoxData[ \(TraditionalForm\`f[x]\)]], ", and ", Cell[BoxData[ \(TraditionalForm\`f[x - h]\)]], ". Also determine the error term. ", StyleBox["Hint", FontSlant->"Italic"], ": for a first derivative, the first power of ", Cell[BoxData[ \(TraditionalForm\`h\)]], " appears in the denominator. What power should be in the denominator for \ a second derivative?" }], "NumberedExercise"], Cell[CellGroupData[{ Cell[BoxData[ \(TraditionalForm\`\((\[ScriptL] = \(\[Alpha]\ \(f(x)\) + \[Beta]\ \(f(x \ - h)\) + \[Gamma]\ \(f(x + h)\)\)\/h\^2)\) + O(h)\)], "Input"], Cell[BoxData[ FormBox[ InterpretationBox[ RowBox[{\(\(\[Alpha]\ \(f(x)\) + \[Beta]\ \(f(x)\) + \[Gamma]\ \(f( x)\)\)\/h\^2\), "+", FractionBox[ RowBox[{ RowBox[{"\[Gamma]", " ", RowBox[{ SuperscriptBox["f", "\[Prime]", MultilineFunction->None], "(", "x", ")"}]}], "-", RowBox[{"\[Beta]", " ", RowBox[{ SuperscriptBox["f", "\[Prime]", MultilineFunction->None], "(", "x", ")"}]}]}], "h"], "+", RowBox[{"(", RowBox[{ RowBox[{\(1\/2\), " ", "\[Beta]", " ", RowBox[{ SuperscriptBox["f", "\[Prime]\[Prime]", MultilineFunction->None], "(", "x", ")"}]}], "+", RowBox[{\(1\/2\), " ", "\[Gamma]", " ", RowBox[{ SuperscriptBox["f", "\[Prime]\[Prime]", MultilineFunction->None], "(", "x", ")"}]}]}], ")"}], "+", InterpretationBox[\(O(h\^1)\), SeriesData[ h, 0, {}, -2, 1, 1]]}], SeriesData[ h, 0, { Plus[ Times[ \[Alpha], f[ x]], Times[ \[Beta], f[ x]], Times[ \[Gamma], f[ x]]], Plus[ Times[ -1, \[Beta], Derivative[ 1][ f][ x]], Times[ \[Gamma], Derivative[ 1][ f][ x]]], Plus[ Times[ Rational[ 1, 2], \[Beta], Derivative[ 2][ f][ x]], Times[ Rational[ 1, 2], \[Gamma], Derivative[ 2][ f][ x]]]}, -2, 1, 1]], TraditionalForm]], "Output"] }, Open ]], Cell[CellGroupData[{ Cell[BoxData[ \(TraditionalForm\`Solve[% \[Equal] \(f\^\[DoublePrime]\)( x), {\[Alpha], \[Beta], \[Gamma]}]\)], "Input"], Cell[BoxData[ \(TraditionalForm\`{{\[Alpha] \[Rule] \(-2\), \[Beta] \[Rule] 1, \[Gamma] \[Rule] 1}}\)], "Output"] }, Open ]], Cell[CellGroupData[{ Cell[BoxData[ \(TraditionalForm\`\[ScriptL] /. First[%] // Simplify\)], "Input"], Cell[BoxData[ \(TraditionalForm\`\(\(-2\)\ \(f(x)\) + f(x - h) + f(h + x)\)\/h\^2\)], \ "Output"] }, Open ]], Cell["Expanding this into a series we get the error term:", "NumberedAnswer"], Cell[CellGroupData[{ Cell[BoxData[ \(TraditionalForm\`% + \(O(h)\)\^3\)], "Input"], Cell[BoxData[ FormBox[ InterpretationBox[ RowBox[{ RowBox[{ SuperscriptBox["f", "\[Prime]\[Prime]", MultilineFunction->None], "(", "x", ")"}], "+", RowBox[{\(1\/12\), " ", RowBox[{ SuperscriptBox["f", TagBox[\((4)\), Derivative], MultilineFunction->None], "(", "x", ")"}], " ", \(h\^2\)}], "+", InterpretationBox[\(O(h\^3)\), SeriesData[ h, 0, {}, 0, 3, 1]]}], SeriesData[ h, 0, { Derivative[ 2][ f][ x], 0, Times[ Rational[ 1, 12], Derivative[ 4][ f][ x]]}, 0, 3, 1]], TraditionalForm]], "Output"] }, Open ]] }, Open ]], Cell[CellGroupData[{ Cell[TextData[{ "Compute a table of values of ", Cell[BoxData[ \(TraditionalForm\`x\^3\)]], " for ", Cell[BoxData[ \(TraditionalForm\`x\)]], " ranging from 0 to 3 in steps of 0.1. Compute ", Cell[BoxData[ \(TraditionalForm\`\(f\^\[DoublePrime]\)[x]\)]], " numerically. You may find the (list) operations ", StyleBox["RotateLeft", "Input"], " and ", StyleBox["RotateRight", "Input"], " useful. Compare your result with a table of exactly computed second \ derivatives. Explain why the results agree exactly (", StyleBox["i.e.", FontSlant->"Italic"], ", there is no error term). Why you do you need to discard the values at \ each end of the resulting list?" }], "NumberedExercise"], Cell[CellGroupData[{ Cell[BoxData[ \(TraditionalForm\`vals = Table[x\^3, {x, 0, 3, 0.1}]\)], "Input"], Cell[BoxData[ \(TraditionalForm\`{0, 0.0010000000000000002`, 0.008000000000000002`, 0.02700000000000001`, 0.06400000000000002`, 0.125`, 0.21600000000000008`, 0.3430000000000001`, 0.5120000000000001`, 0.7290000000000001`, 1.`, 1.3310000000000004`, 1.7280000000000006`, 2.1970000000000005`, 2.7440000000000007`, 3.375`, 4.096000000000001`, 4.913000000000001`, 5.832000000000001`, 6.859000000000001`, 8.`, 9.261000000000001`, 10.648000000000003`, 12.167000000000003`, 13.824000000000005`, 15.625`, 17.576000000000004`, 19.683000000000003`, 21.952000000000005`, 24.38900000000001`, 27.`}\)], "Output"] }, Open ]], Cell[TextData[{ "We compute ", Cell[BoxData[ \(TraditionalForm\`\(f\^\[DoublePrime]\)(x)\)]], " from ", Cell[BoxData[ \(TraditionalForm\`\(\(-2\)\ \(f(x)\) + f(x - h) + f(h + \ x)\)\/h\^2\)]], " using list operations and, apart from the end terms, the agreement is \ excellent. The end terms need to be discarded because these terms involve \ differences of terms at the start and end of the data list." }], "NumberedAnswer"], Cell[CellGroupData[{ Cell[BoxData[ \(TraditionalForm\`\(RotateLeft[vals] + RotateRight[vals] - 2 \ vals\)\/0.1\^2\)], "Input"], Cell[BoxData[ \(TraditionalForm\`{2700.1`, 0.6000000000000001`, 1.2000000000000006`, 1.8`, 2.399999999999999`, 3.000000000000008`, 3.5999999999999917`, 4.200000000000003`, 4.799999999999981`, 5.399999999999982`, 6.000000000000049`, 6.599999999999983`, 7.199999999999961`, 7.800000000000028`, 8.399999999999961`, 9.000000000000162`, 9.59999999999983`, 10.199999999999852`, 10.800000000000052`, 11.399999999999897`, 12.000000000000098`, 12.60000000000012`, 13.199999999999788`, 13.800000000000166`, 14.399999999999123`, 15.000000000000922`, 15.599999999999879`, 16.1999999999999`, 16.800000000000633`, 17.399999999998524`, \(-2961.0999999999985`\)}\)], "Output"] }, Open ]], Cell[CellGroupData[{ Cell[BoxData[ \(TraditionalForm\`Drop[Rest[%], \(-1\)]\)], "Input"], Cell[BoxData[ \(TraditionalForm\`{0.6000000000000001`, 1.2000000000000006`, 1.8`, 2.399999999999999`, 3.000000000000008`, 3.5999999999999917`, 4.200000000000003`, 4.799999999999981`, 5.399999999999982`, 6.000000000000049`, 6.599999999999983`, 7.199999999999961`, 7.800000000000028`, 8.399999999999961`, 9.000000000000162`, 9.59999999999983`, 10.199999999999852`, 10.800000000000052`, 11.399999999999897`, 12.000000000000098`, 12.60000000000012`, 13.199999999999788`, 13.800000000000166`, 14.399999999999123`, 15.000000000000922`, 15.599999999999879`, 16.1999999999999`, 16.800000000000633`, 17.399999999998524`}\)], "Output"] }, Open ]], Cell[CellGroupData[{ Cell[BoxData[ FormBox[ RowBox[{"Table", "[", RowBox[{ RowBox[{"Evaluate", "[", FractionBox[\(\[PartialD]\^2 x\^3\), \(\[PartialD]x\^2\), MultilineFunction->None], "]"}], ",", \({x, 0, 3, 0.1}\)}], "]"}], TraditionalForm]], "Input"], Cell[BoxData[ \(TraditionalForm\`{0, 0.6000000000000001`, 1.2000000000000002`, 1.8000000000000003`, 2.4000000000000004`, 3.`, 3.6000000000000005`, 4.2`, 4.800000000000001`, 5.4`, 6.`, 6.6000000000000005`, 7.200000000000001`, 7.800000000000001`, 8.4`, 9.`, 9.600000000000001`, 10.200000000000001`, 10.8`, 11.4`, 12.`, 12.600000000000001`, 13.200000000000001`, 13.8`, 14.400000000000002`, 15.`, 15.600000000000001`, 16.200000000000003`, 16.8`, 17.400000000000002`, 18.`}\)], "Output"] }, Open ]] }, Open ]] }, Closed]], Cell[CellGroupData[{ Cell[TextData[{ CounterBox["Title"], ". ", CounterBox["Section"], ". ", CounterBox["Subsection"], "\tEuler's method" }], "Subsection"], Cell[TextData[{ "Consider the completely general first-order differential equation, ", Cell[BoxData[ \(TraditionalForm\`y\^\[Prime] = f(x, y)\)]], ". If you know ", Cell[BoxData[ \(TraditionalForm\`y(x)\)]], " at some initial point, say ", Cell[BoxData[ \(TraditionalForm\`y(x\_0)\ = \ y\_0\)]], ", then you can compute ", Cell[BoxData[ \(TraditionalForm\`f(x\_0, y\_0)\)]], " and the value of ", Cell[BoxData[ \(TraditionalForm\`y(x\_0 + h)\ = \ y\_1\)]], " can be obtained approximately from Taylor's theorem, ", StyleBox["i.e", FontSlant->"Italic"], "., ", Cell[BoxData[ \(TraditionalForm\`y\_1 = \ y\_1\ + \ h\ \(f(x\_0, y\_0)\)\)]], ". In general, ", Cell[BoxData[ \(TraditionalForm \`y\_n = y\_\(n - 1\) + h\ \(f(x\_\(n - 1\), y\_\(n - 1\))\)\)]], " where ", Cell[BoxData[ \(TraditionalForm\`x\_n = n\ h\)]], ". This method is known as ", StyleBox["Euler's method", FontSlant->"Italic"], ". The error in each step is of ", Cell[BoxData[ \(TraditionalForm\`O[h]\^2\)]], "." }], "Text"], Cell[CellGroupData[{ Cell["Define Euler's method using dynamic programming by entering:", "NumberedExercise"], Cell[BoxData[ \(TraditionalForm\`x\_n_ = n\ h; \)], "Input"], Cell[BoxData[ \(TraditionalForm\`y\_n_ := \(y\_n = h\ \(f(x\_n, y\_\(n - 1\))\) + y\_\(n - 1\)\)\)], "Input"] }, Open ]], Cell[TextData[{ "Consider the differential equation ", Cell[BoxData[ \(TraditionalForm\`y\^\[Prime] = \(f(x, y) = x\ y\^2\)\)]], ". " }], "Text"], Cell[CellGroupData[{ Cell[TextData[{ "Enter the differential equation as ", Cell[BoxData[ \(TraditionalForm\`f(x_, y_) = x\ y\^2\)]], ". In what follows fix the step size to be ", Cell[BoxData[ \(TraditionalForm\`h = 0.05\)]], " and choose the initial condition ", Cell[BoxData[ \(TraditionalForm\`y\_0 = 1\)]], ". Compute 30 ", Cell[BoxData[ \(TraditionalForm\`{x\_n, y\_n}\)]], ", pairs for ", Cell[BoxData[ \(TraditionalForm\`n\)]], " ranging from 1 to 30 using ", StyleBox["Table", "Input"], ". Use ", StyleBox["ListPlot", "Input"], " with ", StyleBox["PlotJoined\[Rule]True", "Input"], " to visualize the result and call this ", Cell[BoxData[ \(TraditionalForm\`p(1)\)]], "." }], "NumberedExercise"], Cell[BoxData[ \(TraditionalForm\`\(h = 0.05;\)\)], "Input"], Cell[BoxData[ \(TraditionalForm\`\(y\_0 = 1;\)\)], "Input"], Cell[BoxData[ \(TraditionalForm\`\(f(x_, y_) = x\ y\^2;\)\)], "Input"], Cell[CellGroupData[{ Cell[BoxData[ \(TraditionalForm\`Table[{x\_n, y\_n}, {n, 30}]\)], "Input"], Cell[BoxData[ FormBox[ RowBox[{"(", "\[NoBreak]", GridBox[{ {"0.05`", "1.0025`"}, {"0.1`", "1.00752503125`"}, {"0.15000000000000002`", "1.0151383314144649`"}, {"0.2`", "1.0254433897335342`"}, {"0.25`", "1.0385875665528868`"}, {"0.30000000000000004`", "1.0547675285538605`"}, {"0.35000000000000003`", "1.0742368829914637`"}, {"0.4`", "1.097316580607048`"}, {"0.45`", "1.1244089133637387`"}, {"0.5`", "1.1560162984750344`"}, {"0.55`", "1.1927665747393823`"}, {"0.6000000000000001`", 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